I've used Cardano's formula to determine an exact zero of
cube root 4 - cube root 2
The thing i need to do now is "check that it is indeed a zero of the function", I'm not sure how I'm supposed to do that. Isn't that what the point of finding the exact zero was?
If anyone could help me out here I'd really appreciate it.
The Cardano's formula is for finding one exact zero of a cubic polynomial, I'd type it out here but i don't know how to do all the curt and sqrt symbols etc, I know that this isn't wrong because it's an example question on my worksheet, I'm just not sure how to prove it like it's asking me to because I can't set the curt numbers to zero or sub them in or anything to make sure it's a root
curt 4 - curt 2 is the zero, not the polynomial, i need to know how to prove this
For the first term by term I get:
[4^(1/3)-2^(1/3)]^3 = 4 - 3*4^(2/3)*2^(1/3) + 3*4^(1/3)*2^(2/3) - 2
[4^(1/3)-2^(1/3)]^3 = 2 - 3*4^(2/3)*2^(1/3) + 3*4^(1/3)*2^(2/3)
Now 4 = 2^2 so
[4^(1/3)-2^(1/3)]^3 = 2 - 3*2^(4/3)*2(1/3) + 3*2^(2/3)*2(2/3)
[4^(1/3)-2^(1/3)]^3 = 2 - 3*2^(5/3) + 3*2^(4/3)
So
x^3 + 6x - 2 =
2 - 3*2^(5/3) + 3*2^(4/3) + 6*4^(1/3) - 6*2^(1/3) - 2
= -3*2^(5/3) + 3*2^(4/3) + 6*2^(2/3) - 6*2^(1/3)
and 6 = 3*2 = 3*2^(3/3) so the above is
= -3*2^(5/3) + 3*2^(4/3) + 3*2^(5/3) - 3*2^(4/3) = 0.
-Dan
Actually, i looked it over, could you explain to me real quick what you did to get the first equation
[4^(1/3)-2^(1/3)]^3 = 4 - 3*4^(2/3)*2^(1/3) + 3*4^(1/3)*2^(2/3) - 2
I know that ^1/3 is the same as a cube root, but i don't get why you cubed the left side, and i don't see what happened to get the right side